Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  double s, a, b;
  cin >> s >> a >> b;
  int n;
  cin >> n;
  vector<int> segmentsbeg;
  vector<int> segmentsend;
  segmentsbeg.reserve(n);
  segmentsend.reserve(n);
  int x, y;
  for (int i = 0; i < n; ++i) {
    cin >> x >> y;
    segmentsbeg.emplace_back(x);
    segmentsend.emplace_back(y);
  }
  vector<int> prefix;
  prefix.reserve(n + 1);
  prefix.emplace_back(0);
  for (int i = 1; i < n + 1; ++i) {
    prefix.emplace_back(prefix[i - 1] + segmentsend[i - 1] -
                        segmentsbeg[i - 1]);
  }
  vector<double> ans;
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> x >> y;
    double v = y / (y - s);
    double xbeg1 = a + (x - a) * (-s) / (y - s);
    double xbeg2 = b + (x - b) * (-s) / (y - s);
    int beg1 = static_cast<int>(xbeg1);
    int beg2 = static_cast<int>(xbeg2);
    int ind1 = upper_bound(segmentsbeg.begin(), segmentsbeg.end(), beg1) -
               segmentsbeg.begin();
    int ind2 = upper_bound(segmentsend.begin(), segmentsend.end(), beg2) -
               segmentsend.begin();
    --ind1;
    double dist = 0;
    if (ind1 == ind2) {
      cout << (xbeg2 - xbeg1) / v << endl;
    } else {
      if (ind1 != -1 && static_cast<double>(segmentsend[ind1]) > xbeg1) {
        dist += static_cast<double>(segmentsend[ind1]) - xbeg1;
      }
      if (ind2 != n && static_cast<double>(segmentsbeg[ind2]) < xbeg2) {
        dist += xbeg2 - static_cast<double>(segmentsbeg[ind2]);
      }
      ++ind1;
      dist += prefix[ind2] - prefix[ind1];
      cout << dist / v << endl;
    }
  }
  return 0;
}